Question

Let \[ M = \begin{pmatrix} 2 & 0 & -1 \\ 4 & 1 & -4 \\ 2 & 0 & x \end{pmatrix} \] for some real number \( x \).
If \( 0 \) is an eigenvalue of \( M \), then \[ (M^4 + M) \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \] is equal to

• A. \( \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \)
• B. \( \begin{pmatrix} 2 \\ 0 \\ 2 \end{pmatrix} \)
• C. \( \begin{pmatrix} 5 \\ 0 \\ 5 \end{pmatrix} \)
• D. \( \begin{pmatrix} 17 \\ 0 \\ 17 \end{pmatrix} \)

Hint:

When an eigenvalue is 0, the matrix will annihilate the corresponding eigenvector when multiplied. Thus, powers of the matrix will also annihilate the eigenvector.

Answer 1

The Correct Option is B

Step 1: Use the eigenvalue condition.
Since 0 is an eigenvalue of \( M \), there exists a nonzero vector \( v \) such that: \[ M v = 0 v = 0 \] Let the eigenvector corresponding to eigenvalue 0 be \( v = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \).

Step 2: Compute \( (M^4 + M)v \).
Using the fact that \( Mv = 0 \), it follows that: \[ M^2 v = M(Mv) = M(0) = 0 \] Thus, \[ M^4 v = 0 \quad \text{and} \quad Mv = 0 \] Therefore, \[ (M^4 + M) v = 0 + 0 = 0 \]

Final Answer:
\[ \boxed{\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}} \]